A `4 muF` capacitor and a resistance of `2.5 M Omega` are series 12 V battery. Find the time after with the potential difference across the capacitor is 3 times the potential difference across the resistance [Given `ln(2) = 0.693`]

Options

A. `13.86 s`
B. `6.93 s`
C. `7 s`
D. `14 s`

Answer

A

Explanation

a. Let at any instant of time t during charging process, the
transiet current in the circuit be I. `I=(V_0//R)e^(-t//RC)`. Therefore,
potential difference across resistor R is
`[(V_0/R)e^(-t//RC)] xx R = V_0 e^(-t//RC)`
Therefore, potential difference across C is
`V_0-V_0e^(-t//RC) = V_0(1-e^(-t//RC))`
or `1-e^(-t//RC) = 3e^(-t//RC) or 1=4e^(-t//RC)`
Taking log on both sides, we get
`log_e1 = 2log_e 2+(-t/RC)`
or `0=2 xx 2.303 log_(10) 2-t/(RC)`
or `t = (2xx2.303log_(10)2) xx 2.5 xx 10^(6) xx 4 xx 10 = 13.86 s`,